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Home/ Questions/Q 6730155
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Editorial Team
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Editorial Team
Asked: May 26, 20262026-05-26T10:22:26+00:00 2026-05-26T10:22:26+00:00

I would like to write a function that has a loop in it which

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I would like to write a function that has a loop in it which preforms the operations necessary for Muller’s method.

f[x_] := x^3 - x - 1;
x0 = 0.8
x1 = 1.5
x2 = 2.0
x3 = 5.0;
\[Epsilon] = 0.001;

While[(Abs[f[x3]] >= \[Epsilon]),
 h0 = x1 - x0;
 h1 = x2 - x1;
 d0 = (f[x1] - f[x0])/h0;
 d1 = (f[x2] - f[x1])/h1;
 A = (d1 - d0)/(h1 + h0);
 B = A*h1 + d1;
 Cx = f[x2];
 raiz = Sqrt[B^2 - 4.0*A*Cx];
 If[Abs[B + raiz] > Abs[B - raiz], dens = B + raiz, dens = B - raiz];
 x3 = (x2 - 2*Cx)/dens;
 i++;
 Print["Iteration: ", i, "\t root \[TildeTilde] ", x3];
 x0 = x1;
 x1 = x2;
 x2 = x3;
 ]

But I get infinit loop…

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-26T10:22:27+00:00Added an answer on May 26, 2026 at 10:22 am

    Muller’s method following Eric (Always better than Wikipedia):
    Thanks to Heike for pointing out a few errors in the comment below

    h[x_] := HermiteH[24, x];
    i = Length@CoefficientList[h[x], x] - 1;
    f[i, x_] := h[x];
    roots = {};
    While[ i > 1,
      x0 = -2; x1 = -1; x2 = -.5; k = 1;
      While[Abs[k] > .001,
       q = (x0 - x1)/(x1 - x2);
       a = q f[i, x0] - q (1 + q) f[i, x1] + q^2 f[i, x2];
       b = (2 q + 1) f[i, x0] - (1 + q)^2 f[i, x1] + q^2 f[i, x2];
       c = (1 + q) f[i, x0];
       p = Sqrt[b b - 4 a c];
       xp = x0 - (x0 - x1) 2 c /(k = If[Abs[b + p] > Abs[b - p], b + p, b - p]);
       {x2, x1, x0} = {x1, x0, xp};
       ];
      AppendTo[roots, xp];
      i--;
      f[i, x_] = f[i + 1, x]/(x - xp);
      ];
    Show[
     Plot[h[x], {x, -2, 2}],
     Graphics[{PointSize[Large], Point[{#, 0} & /@ roots]}]]
    

    enter image description here

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